Prove that the following matrix is positive definite. $$ A = \begin{bmatrix} 1 & \frac12 & \dots & \frac1n \\ \frac12 & \frac13 & \dots & \frac1{n+1} \\ \vdots & \vdots & \ddots & \vdots \\ \frac1n & \frac1{n+1} & \dots & \frac1{2n-1} \end{bmatrix} $$
I tried to do a couple of thing without succes (it's immediate that $A=A^t$):
First: I tried to show this by definition: In this case, the matrix is real so, let $v^t=(x_1,...,x_n)^t$ be a real vector, I tried to show that $v^tAv>0$. I then arrive to $$x_1^2+x_1x_2+\frac{2x_1x_3}{3}+\dots+\frac{x_2x_n}{n+1}+\frac{x_n^2}{2n-1}$$ but I don't know how to show that this last equation is $>0$, is there some identity to show this?
Second, I tried to define the matrix as a inner product, that is $a_{ij} = \langle v_j, v_i \rangle$ where $\{v_1,\dots,v_n\}$ is a ortonormal base for $\mathbb{R}^n$. How can I define those $v_j$'s?
Third, I tried to show that the minors of the matrix are all positives, I tried to do it by induction over the matrix size, for $n=1$ is trivial because $A$ is $1>0$ in this case, for $n=2$, we have
$$A=\begin{bmatrix} 1 & 1/2 \\ 1/2 & 1/3 \\ \end{bmatrix}$$
and then
$$\begin{vmatrix} 1 & 1/2 \\ 1/2 & 1/3 \\ \end{vmatrix}=1/3-1/4>0$$
If we suposse that for size $n-1$ we have determinant positive then I tried to calculate $\det(A)$ using determinants of size $n-1$ all positives. (I did it expanding by the first column):
$$\det(A)=1\det\begin{bmatrix} 1/3 & \dots & 1/n+1 \\ \vdots & \vdots & \vdots \\ 1/n+1 & \dots & 1/2n-1 \\ \end{bmatrix}-\frac{1}{2}\det\begin{bmatrix} 1/2 & \dots & 1/n \\ \vdots & \vdots & \vdots \\ 1/n+1 & \dots & 1/2n-1 \\ \end{bmatrix}+\dots+(-1)^n\frac{1}{n}\det\begin{bmatrix} 1/2 & \dots & 1/n \\ \vdots & \ddots & \vdots \\ 1/n & \dots & 1/2n-2 \\ \end{bmatrix}$$
Then, how can I do to show $\det(A) > 0$ keeping in mind that every determinant is positive?
